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\title{\heiti\zihao{2} 习题5.2}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{使用 Lagrange 余项估计下列近似公式的误差 :}
\subsection{$\ln (1+x) \approx x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{6},$ 当 $0 \leqslant x \leqslant 1$;}
\textbf{解}\quad
$\begin{array}{c}\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{6}+\frac{x^{7}}{7(1+\xi x)^{7}}, \quad 0 \leqslant \xi \leqslant 1 \\R_{n}(x)=\left|\frac{x^{7}}{7(1+\xi x)^{7}}\right| \leqslant \frac{1}{7} \approx 0.1429\end{array}$


\subsection{$\cos x \approx 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720},$ 当 $|x|<1$.}
\textbf{解}\quad
$\begin{array}{l}\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\frac{\cos (\theta x)}{8 !} x^{8} \\R_{n}(x)\left|\frac{\cos (\theta x) x^{8}}{8 !}\right| \leqslant \frac{x^{8}}{8 !} \leqslant \frac{1}{8 !} \approx 2.42 \times 10^{-5}\end{array}$

\section{写出下列函数的带 $\mathrm{Peano}$ 余项的$\mathrm{Maclaurin}$ 公式:}
\subsection{$\mathrm{e}^{a x}(a \neq 0)$}
\textbf{解}\quad
$a_k=\frac{f^{(k)}(0)}{k!}=\frac{a^k}{k!}$,所以$e^{ax}=1+\frac{a}{1}x+\frac{a^2}{2!}x^2+\cdots+\frac{a^n}{n!}x^n+o(x^n)$


\subsection{$\sin x^{2}$;}
\textbf{解}\quad
将$x^2$视为$u$,则$\sin x^2=x^{2}-\frac{x^{6}}{3 !}+\frac{x^{10}}{5 !}-\frac{x^{14}}{7 !}+\cdots+\frac{(-1)^{n-1}}{(2 n-1) !} x^{4 n-2}+o\left(x^{4 n}\right)$


\subsection{$\ln (1-x)$;}
\textbf{解}\quad
$\ln(1-x)=-\left[x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{1}{n} x^{n}+o\left(x^{n}\right)\right]$


\subsection{$\frac{1}{(1+x)^{2}}$}
\textbf{解}\quad
已知 $(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right),$ 当$\alpha=-2$ 时$\frac{1}{(1+x)^{2}}=1-2 x+\frac{-2(-2-1)}{2} x^{2}+\cdots+\frac{-2(-2-1) \cdots(-2-n+1)}{n !} x^{n}+o\left(x^{n}\right)\quad=1-2 x+3 x^{2}+\cdots+(-1)^{n}(n+1) x^{n}+o\left(x^{n}\right)$


\subsection{$\frac{x^{3}+2 x+1}{x+1}$}
\textbf{解}\quad
$\frac{x^{3}+2 x+1}{x+1}=x^{2}-x+3-2 \frac{1}{1+x},$ 又已知$$\frac{1}{1+x}=1-x+x^{2}-x^{3}+\cdots+(-1)^{n} x^{n}+o\left(x^{n}\right), \quad x \rightarrow 0$$

所以$$\begin{aligned}\frac{x^{3}+2 x+1}{x+1} &=x^{2}-x+3-2\left[1-x+x^{2}-x^{3}+\cdots+(-1)^{n} x^{n}+o\left(x^{n}\right)\right] \\&=1+x-x^{2}+2 x^{3}+\cdots+(-1)^{n+1} 2 x^{n}+o\left(x^{n}\right)\end{aligned}$$


\subsection{$\cos ^{3} x+\sin ^{3} x$}
\textbf{解}\quad
$\sin 3\alpha=3\sin\alpha-4\sin^3\alpha,\cos 3\alpha=-3\cos\alpha+4\cos^3\alpha$从而可得$\sin ^{3} \alpha+\cos ^{3} \alpha=\frac{3}{4} \sin \alpha-\frac{1}{4} \sin 3\alpha+\frac{3}{4} \cos \alpha+\frac{1}{4} \cos 3 \alpha$将上述式子泰勒展开即可.


\subsection{$\ln \frac{1+x}{1-2 x}$}
\textbf{解}\quad
解 $\ln \frac{1+x}{1-2 x}=\ln (1+x)-\ln (1-2 x),$ 已知$$\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{(-1)^{n-1}}{n} x^{n}+o\left(x^{n}\right)$$

所以$$\begin{aligned}\ln (1-2 x) &=-2 x-\frac{(2 x)^{2}}{2}-\frac{(2 x)^{3}}{3}-\cdots-\frac{1}{n}(2 x)^{n}+o\left(x^{n}\right) \\&=-\left[2 x+\frac{(2 x)^{2}}{2}+\frac{(2 x)^{3}}{3}+\cdots+\frac{1}{n}(2 x)^{n}+o\left(x^{n}\right)\right] \\&=-\left[2 x+\frac{2^{2}}{2} x^{2}+\frac{2^{3}}{3} x^{3}+\cdots+\frac{2^{n}}{n} x^{n}+o\left(x^{n}\right)\right]\end{aligned}$$

从而可知$$\begin{aligned}\ln \frac{1+x}{1-2 x}=& \ln (1+x)-\ln (1-2 x) \\=& x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{(-1)^{n-1}}{n} x^{n}+o\left(x^{n}\right)-\left\{-\left[2 x+\frac{2^{2}}{2} x^{2}+\frac{2^{3}}{3} x^{3}+\cdots+\frac{2^{n}}{n} x^{n}+o\left(x^{n}\right)\right]\right\} \\=& x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{(-1)^{n-1}}{n} x^{n}+\left(2 x+2 x^{2}+\frac{2^{3}}{3} x^{3}+\cdots+\frac{2^{n}}{n} x^{n}\right)+o\left(x^{n}\right) \\=& 3 x-\frac{3 x^{2}}{2}+\frac{7 x^{3}}{3}+\cdots+\frac{2^{n}+(-1)^{n-1}}{n} x^{n}+o\left(x^{n}\right)\end{aligned}$$

\section{求出 $\arcsin x$ 的带 Peano 余项的 Maclaurin 公式.}
\textbf{解}\quad
$$(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-\frac{1}{2}}$$

$$(1+u)^{-\frac{1}{2}}=1+\sum_{k=1}^{n} \frac{-\frac{1}{2}\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 k-1}{2}\right)}{k !} u^{k}+o(u^{k})$$

令$u=-x^2$可得$$(1-x^2)^{-\frac{1}{2}}=1+\sum_{k=1}^{n} \frac{\frac{1}{2}\left(\frac{3}{2}\right) \cdots\left(\frac{2 k-1}{2}\right)}{k !} x^{2k}+o(x^{2k})$$

所以$$\arcsin x=x+\sum_{k=1}^{n} \frac{\frac{1}{2}\left(\frac{3}{2}\right) \cdots\left(\frac{2 k-1}{2}\right)}{k !(2k+1)} x^{2k+1}$$

\section{按指定要求写出下列函数带 $\mathrm{Peano}$ 余项的 $\mathrm{Maclaurin}$ 公式:}
\subsection{$\mathrm{e}^{x} \cos x$ 到含 $o\left(x^{4}\right)$ 的项 }
\textbf{解}\quad
$\quad \mathrm{e}^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+o\left(x^{4}\right), \cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+o\left(x^{4}\right),$ 所以
$$\begin{aligned} \mathrm{e}^{x} \cos x &=\left[1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+o\left(x^{4}\right)\right]\left[1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+o\left(x^{4}\right)\right] \\ &=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}-\frac{x^{2}}{2 !}-\frac{x^{3}}{2 !}-\frac{x^{4}}{4}+\frac{x^{4}}{4 !}+o\left(x^{4}\right) \\ &=1+x-\frac{x^{3}}{3}+\frac{x^{4}}{6}+o\left(x^{4}\right) \end{aligned}$$


\subsection{$\frac{x}{\sin x}$ 到含 $o\left(x^{4}\right)$ 的项 }
\textbf{解}\quad
$$\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+o\left(x^{5}\right)$$

$$\frac{x}{\sin x}=\frac{1}{1-\frac{x^{2}}{3 !}+\frac{x^{4}}{5 !}+o\left(x^{4}\right)}=\frac{1}{1-\left[\frac{x^{2}}{3 !}-\frac{x^{4}}{5 !}+o\left(x^{4}\right)\right]}$$

$$
	\Delta=\frac{x^{2}}{3 !}-\frac{x^{4}}{5 !}+o\left(x^{4}\right), \quad \Delta^{2}=\frac{x^{4}}{36}+o\left(x^{4}\right), \quad \Delta^{3}=o\left(x^{4}\right)
$$
$$
	\begin{aligned}
		\frac{x}{\sin x} & =\frac{1}{1-\Delta}=\frac{1+\Delta+\Delta^{2}}{1-\Delta^{3}}                                            \\
		                 & =\frac{1+\frac{x^{2}}{6}+\frac{x^{4}}{5 !}+\frac{x^{4}}{36}+o\left(x^{4}\right)}{1+o\left(x^{4}\right)} \\&=1+\frac{x^{2}}{6}+\frac{13}{360} x^{4}+o\left(x^{4}\right)
	\end{aligned}
$$



\subsection{$\frac{x}{2 x^{2}+x-1}$ 到含 $o\left(x^{4}\right)$ 的项 }
\textbf{解}\quad
$$
	\begin{aligned}
		\frac{x}{2 x^{2}+x-1} & =\frac{1}{3}\left(\frac{1}{x+1}+\frac{1}{2 x-1}\right)=\frac{1}{3}\left(\frac{1}{x+1}-\frac{1}{1-2 x}\right) \\
		\frac{1}{1+x}         & =1-x+x^{2}-x^{3}+x^{4}+o\left(x^{4}\right)                                                                   \\
		\frac{1}{1-2 x}       & =1+2 x+(2 x)^{2}+(2 x)^{3}+(2 x)^{4}+o\left(x^{4}\right)                                                     \\
		                      & =1+2 x+4 x^{2}+8 x^{3}+16 x^{4}+o\left(x^{4}\right)
	\end{aligned}
$$
$$
	\begin{aligned}
		\frac{x}{2 x^{2}+x-1} & =\frac{1}{3}\left[-3 x-3 x^{2}-9 x^{3}-15 x^{4}+o\left(x^{4}\right)\right] \\&=-x-x^{2}-3 x^{3}-5 x^{4}+o\left(x^{4}\right)
	\end{aligned}
$$

\subsection{$\mathrm{e}^{x-x^{2}}$ 到含 $o\left(x^{5}\right)$ 的项}
\textbf{解}\quad

$\begin{aligned}\mathrm{e}^{x-x^{2}} &=1+\left(x-x^{2}\right)+\frac{\left(x-x^{2}\right)^{2}}{2 !}+\frac{\left(x-x^{2}\right)^{3}}{3 !}+\frac{\left(x-x^{2}\right)^{4}}{4 !}+\frac{\left(x-x^{2}\right)^{5}}{5 !}+o\left(x^{5}\right) \\&=1+\left(x-x^{2}\right)+\frac{x^{2}-2 x^{3}+x^{4}}{2 !}+\frac{\left(x-x^{2}\right)\left(x^{2}-2 x^{3}+x^{4}\right)}{3 !}+\frac{\left(x^{2}-2 x^{3}+x^{4}\right)^{2}}{4 !}+\frac{\left(x-x^{2}\right)^{5}}{5 !}+o\left(x^{5}\right) \\&=1+\left(x-x^{2}\right)+\frac{x^{2}-2 x^{3}+x^{4}}{2 !}+\frac{x^{3}-3 x^{4}+3 x^{5}}{3 !}+\frac{x^{4}+2 x^{5}}{4 !}+\frac{x^{2}}{5 !}+o\left(x^{5}\right) \\&=1+x-\frac{59 x^{2}}{120}-\frac{5 x^{3}}{6}+\frac{x^{4}}{24}+\frac{7 x^{5}}{12}+o\left(x^{5}\right)\end{aligned}$


\subsection{$\sqrt{1-3 x+x^{3}}-\sqrt[3]{1-2 x+x^{2}}$ 到含 $o\left(x^{3}\right)$ 的项}
\textbf{解}\quad
已知$$\begin{aligned}(1+x)^{a}=& 1+\alpha x+\frac{\alpha(\alpha-1)}{2} x^{2}+\frac{\alpha(\alpha-1)(\alpha-2)}{3 !} x^{3}+\cdots+ \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right)\end{aligned}$$

$$\begin{aligned}(1+x)^{\frac{1}{2}}&=1+\frac{x}{2}-\frac{1}{2 \cdot 4} x^{2}+\frac{3}{6 ! !} x^{3}-\frac{5 ! !}{8 ! !} x^{4}+o\left(x^{4}\right) \\&=1+\frac{x}{2}-\frac{1}{8} x^{2}+\frac{1}{16} x^{3}-\frac{15}{384} x^{4}+o\left(x^{4}\right) \\(1+x)^{\frac{2}{3}}&= 1+\frac{2 x}{3}+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2} x^{2}+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{3 !} x^{3} +\frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right)}{4 !} x^{4}\\&- \frac{2 x}{3}-\frac{1}{9} x^{2}+\frac{4}{81} x^{3}-\frac{28}{24 \cdot 3^{4}} x^{4}+o\left(x^{4}\right) \\&= 1+\frac{2 x}{3}-\frac{1}{9} x^{2}+\frac{4}{81} x^{3}-\frac{7}{486} x^{4}+o\left(x^{4}\right)\end{aligned}$$

所以分别将$x^3-3x$和$-x$带入上面两个式子中,可得原式=$-\frac{5}{6}x-\frac{73}{72}x^2-\frac{1475}{1296}x^3+o(x^3)$


\subsection{$\sqrt[3]{\sin x^{3}}$ 到含 $x^{13}$ 的项}
\textbf{解}\quad

$$
	\begin{aligned}\sqrt[3]{\sin x^{3}} & =\left[x^{3}-\frac{x^{9}}{3 !}+\frac{x^{15}}{5 !}+o\left(x^{15}\right)\right]^{\frac{1}{3}}=x\left[1+\left(\frac{x^{12}}{120}-\frac{x^{6}}{6}+o\left(x^{12}\right)\right)\right]^{\frac{1}{3}} \\&=x\left[1+\frac{1}{3}\left(\frac{x^{12}}{120}-\frac{x^{6}}{6}+o\left(x^{12}\right)\right)-\frac{1}{9}\left(\frac{x^{12}}{120}-\frac{x^{6}}{6}+o\left(x^{12}\right)\right)^{2}+o\left(x^{12}\right)\right] \\&=x-\frac{7}{18} x^{7}-\frac{1}{3240} x^{13}+o\left(x^{13}\right)
	\end{aligned}
$$


\subsection{$\ln (\cos x+\sin x)$ 到含 $o\left(x^{4}\right)$ 的项}
\textbf{解}\quad

$\begin{array}{l}\sin x=x-\frac{x^{3}}{3 !}+o\left(x^{4}\right) \\\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+o(x)^{4} \\\sin x+\cos x=1+x-\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+o\left(x^{4}\right)\\\ln (1+t)=t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\frac{t^{4}}{4}+o\left(t^{4}\right)\end{array}$
$$
	\begin{array}{l}t=x-\frac{x^{2}}{2}-\frac{x^{3}}{6}+\frac{x^{4}}{24} \\\ln (\sin x+\cos x)=x-x^{2}+\frac{2}{3} x^{3}-\frac{2}{3} x^{4}+o\left(x^{4}\right)\end{array}
$$

\section{写出下列函数在指定点的带Peano余项的Taylor公式:}
\subsection{$f(x)=\ln x, x_{0}=2$}
\textbf{解}\quad
$$a_{k}=\frac{f^{(k)}(2)}{k !}=\frac{2^{-k}(-1)^{k-1}(k-1) !}{k !}=\frac{(-1)^{k-1}}{2^{k} k}$$

所以$$f(x)=\ln 2+\sum_{k=1}^{n} \frac{(-1)^{k-1}}{2^{k} k}(x-2)^{k}+o\left((x-2)^{n}\right)$$

\subsection{$f(x)=\sin x, x_{0}=1$}
\textbf{解}\quad
$f(x)=\sin x, f^{(n)}\left(x_{0}\right)=\sin \left(x_{0}+\frac{n \pi}{2}\right)$

$$\begin{aligned}f(x)&= f(1)+f^{\prime}(1)(x-1)+\frac{f^{\prime \prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime \prime}(1)}{3 !}(x-1)^{3}+\cdots+\\& \frac{f^{(n)}(1)}{n !}(x-1)^{n}+o\left((x-1)^{\prime \prime}\right) \\&= \sin 1+\cos 1 \cdot(x-1)-\frac{\sin 1}{2 !}(x-1)^{2}-\frac{\cos 1}{3 !}(x-1)^{3}+\\& \frac{\sin 1}{4 !}(x-1)^{4}+\cdots+\frac{1}{n !} \sin \left(1+\frac{n \pi}{2}\right)(x-1)^{n}+o\left((x-1)^{n}\right)\end{aligned}$$


\subsection{$f(x)=e^{x},x_{0}=1$}
\textbf{解}\quad
$$f(x)=\mathrm{e} \cdot \mathrm{e}^{x-1}=\mathrm{e}+(x-1) \mathrm{e}+\frac{(x-1)^{2}}{2 !} \mathrm{e}+\frac{(x-1)^{3}}{3 !} \mathrm{e}+\cdots+\frac{(x-1)^{n}}{n !} \mathrm{e}+o\left((x-1)^{n}\right)$$




\section{利用 $\mathrm{Taylor}$ 公式,求下列极限 :}
\subsection{$\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x^{3}}-1-x^{3}}{\sin ^{6} 2 x}$}
\textbf{解}\quad
分母 $\sin ^{6} 2 x \sim 2^{6} x^{6},(x \rightarrow 0),$ 所以分子中的函数需要展开到 $x^{6}$,
$$
	\mathrm{e}^{x^{3}}=1+x^{3}+\frac{x^{6}}{2 !}+o\left(x^{6}\right)
$$

所以
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x^{3}}-1-x^{3}}{\sin ^{6} 2 x} & =\lim _{x \rightarrow 0} \frac{1+x^{3}+\frac{x^{6}}{2}+o\left(x^{6}\right)-1-x^{3}}{2^{6} x^{6}} \\
		                                                                         & =\lim _{x \rightarrow 0} \frac{\frac{x^{6}}{2}+o\left(x^{6}\right)}{2^{6} x^{6}}=\frac{1}{2^{7}}
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow+\infty}\left[\sqrt[3]{x^{3}+3 x}-\sqrt{x^{2}-2 x}\right]$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow+\infty}\left[\sqrt[3]{x^{3}+3 x}-\sqrt{x^{2}-2 x}\right] & =\lim _{x \rightarrow+\infty}\left(\sqrt[3]{1+3 \frac{1}{x^{2}}}-\sqrt{1-\frac{2}{x}}\right) \cdot x \\&=\lim _{t \rightarrow 0^{+}} \frac{\left(\sqrt[3]{1+3 t^{2}}-\sqrt{1-2 t}\right)}{t}\\&=\lim _{t \rightarrow 0^{+}} \frac{1+\frac{t^{2}}{3}-1+t+o (t^{2})}{t}\\&=1
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow+\infty} x^{\frac{3}{2}}[\sqrt{x+1}+\sqrt{x-1}-2 \sqrt{x}]$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow+\infty} x^{\frac{3}{2}}[\sqrt{x+1}+\sqrt{x-1}-2 \sqrt{x}] & =\lim _{x \rightarrow+\infty} x^{2}\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}-2\right) \\&=\lim _{t \rightarrow 0^{+}} \frac{\sqrt{1+t}+\sqrt{1-t}-2}{t^{2}}\\&=\lim _{t+0^{+}} \frac{-\frac{t^{2}}{8}-\frac{t^{2}}{8}}{t^{2}}\\&=-\frac{1}{4}
	\end{aligned}
$$


\subsection{$\lim _{x \rightarrow 0} \frac{e^{x} \sin x-x-x^{2}}{x \sin x \arcsin x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{e^{x} \sin x-x-x^{2}}{x \sin x \arcsin x} & =\lim _{x \rightarrow 0} \frac{e^{x} \sin x-x-x^{2}}{x^3} \\&=\lim _{x \rightarrow 0} \frac{\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}\right)\left(x-\frac{x^{3}}{6}\right)-x-x^2}{x^{3}}\\&=\frac{1}{3 }
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\sqrt{1+2 \sin x}-\mathrm{e}^{x}+x^{2}}{x^{3}}$}
\textbf{解}\quad
$$
	\begin{array}{c}
		(1+2 \sin x)^{\frac{1}{2}}=1+\frac{2 x}{2}-\frac{1}{8}(2 x)^{2}+\frac{1}{16}(2 x)^{3}+o\left(x^{3}\right) \\
		e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+o\left(x^{3}\right)
	\end{array}
$$
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{\sqrt{1+2 \sin x}-\mathrm{e}^{x}+x^{2}}{x^{3}} & =\lim _{x \rightarrow 0} \frac{-\frac{x^{3}}{6}+\frac{x^{3}}{2}-\frac{x^{3}}{6}+o\left(x^{3}\right)}{x^{3}} \\
		                                                                             & =\lim _{x \rightarrow 0} \frac{\frac{x^{3}}{6}+o\left(x^{3}\right)}{x^{3}}=\frac{1}{6}
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{1-(\cos x)^{\sin x}}{x^{3}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		(\cos x)^{\sin x} & =\mathrm{e}^{\sin x \ln (\cos x)}=\mathrm{e}^{\left[x-\frac{x^{3}}{6}+o\left(x^{3}\right)\right] \ln \left[1-\frac{x^{2}}{2}+o\left(x^{2}\right)\right]}                                                              \\
		                  & =\mathrm{e}^{\left[x-\frac{x^{3}}{6}+o\left(x^{3}\right)\right]\left\{\left[-\frac{x^{2}}{2}+o\left(x^{3}\right)\right]-\frac{1}{2}\left[-\frac{x^{2}}{2}+o\left(x^{3}\right)\right]^{2}+o\left(x^{3}\right)\right\}} \\
		                  & =\mathrm{e}^{\left[x-\frac{x^{3}}{6}+o\left(x^{3}\right)\right]\left[-\frac{x^{2}}{2}+o\left(x^{3}\right)\right]}=\mathrm{e}^{-\frac{x^{3}}{2}+o\left(x^{3}\right)}                                                   \\
		                  & =1+\left[-\frac{x^{3}}{2}+o\left(x^{3}\right)\right]+\frac{1}{2}\left[-\frac{x^{3}}{2}+o\left(x^{3}\right)\right]^{2}+o\left(x^{3}\right)                                                                             \\
		                  & =1-\frac{x^{3}}{2}+o\left(x^{3}\right)
	\end{aligned}
$$

$$\lim _{x \rightarrow 0} \frac{1-(\cos x)^{\sin x}}{x^{3}}=\frac{1}{2}$$

\section{利用 Taylor 公式,求下列数列极限 :}
\subsection{$\lim _{n \rightarrow \infty} n^{2} \ln \left(n \sin \frac{1}{n}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{n \rightarrow \infty} n^{2} \ln \left(n \sin \frac{1}{n}\right) & =\lim _{t \rightarrow 0} \frac{\ln \left(\frac{\sin t}{t}\right)}{t^{2}} \\&=\lim _{t \rightarrow 0} \frac{\frac{t}{\sin t} \cdot \frac{\cos t \cdot t-\sin t}{t^{2}}}{2 t}\\&=\lim _{t \rightarrow 0} \frac{t\left(1-\frac{t^{2}}{2}+o\left(t^{2}\right)\right)-t+\frac{t^{3}}{6}+o(t^3)}{2 t^{3}}\\&=\lim _{t \rightarrow 0} \frac{-\frac{t^{3}}{2}+\frac{t^{3}}{b}}{2 t^{3}}\\&=-\frac{1}{6}
	\end{aligned}
$$


\subsection{$\lim _{n \rightarrow \infty}(-1)^{n} n \sin \left(\sqrt{n^{2}+2} \pi\right)$}
\textbf{解}\quad
$$
	\begin{aligned}\lim _{n \rightarrow \infty}(-1)^{n} n \sin \left(\pi \sqrt{n^{2}+2}\right) &=\lim _{n \rightarrow \infty} n \sin \left[\pi\left(\sqrt{n^{2}+2}-n\right)\right] \\&=\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{\sqrt{n^{2}+2}+n} \\&=\lim _{n \rightarrow \infty} n\left[\frac{2 \pi}{\sqrt{n^{2}+2}+n}-\frac{1}{3}\left(\frac{2 \pi}{\sqrt{n^{2}+2}+n}\right)^{2}+o\left(\frac{1}{n^{2}}\right)\right] \\&=\lim _{n \rightarrow \infty}\left[\frac{2 \pi}{\sqrt{1+\frac{2}{n}}+1}+o\left(\frac{1}{n}\right)\right]\\&=\pi\end{aligned}
$$

\section{设 $f(x)$ 在 $x=0$ 点处存在二阶导数,且有$\lim _{x \rightarrow 0}\left(1+\frac{f(x)}{x}\right)^{\frac{1}{x}}=\mathrm{e}^{3}$}
\subsection{求 $f(0), f^{\prime}(0), f^{\prime \prime}(0)$}
\textbf{解}\quad
显然$f(0)$是0.
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{\ln(1+\frac{f(x)}{x})}{x} & =\lim _{x\rightarrow 0} \frac{f(x)}{x^{2}} \\&=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{2x}\\&=\lim _{x \rightarrow 0} \frac{f^{\prime\prime}(x)}{2}\\&=3
	\end{aligned}
$$

由此可得$$f^{\prime}(0)=0,f^{\prime\prime}(0)=6$$


\subsection{求 $\lim _{x \rightarrow 0}\left(1+\frac{f(x)}{x}-x\right)^{\frac{1}{x}}$.}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow 0}\frac{ \ln(1+\frac{f(x)}{x}-x)}{x} & =\lim _{x \rightarrow 0}\frac{f(x)-x^2}{x^2} \\&=\lim _{x \rightarrow 0} \frac{0+0+\frac{f^{\prime \prime}(0)}{2 !} x^{2}-x^{2}}{x^{2}}\\&=2
	\end{aligned}
$$

原式极限为$e^2$

\section{当 $x>0$ 时,求证 : 对任何 $n \in \mathrm{N}^{*}$,有$x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots-\frac{x^{2 n}}{2 n}<\ln (1+x)<x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots+\frac{x^{2 n-1}}{2 n-1}$}
\textbf{证}\quad
分别将$\ln(1+x)$用带拉格朗日余项的泰勒公式展开到第$2n,2n+1$项,有
$$
	\begin{array}{l}
		\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots+\frac{x^{2 n-1}}{2 n-1}-\frac{x^{2 n}}{2 n\left(1+\xi_{1}\right)^{2 n}}, \quad 0<\xi_{1}<x \\
		\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots-\frac{x^{2 n}}{2 n}+\frac{x^{2 n+1}}{(2 n+1)\left(1+\xi_{2}\right)^{2 n+1}}, \quad 0<\xi_{2}<x
	\end{array}
$$

显然当 $x>0$ 时 $, \frac{x^{2 n}}{2 n\left(1+\xi_{1}\right)^{2 n}}>0, \frac{x^{2 n+1}}{(2 n+1)\left(1+\xi_{2}\right)^{2 n+1}}>0,$ 所以
$$
	x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots-\frac{x^{2 n}}{2 n}<\ln (1+x)<x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots+\frac{x^{2 n-1}}{2 n-1}
$$

得证.

\section{使用 Lagrange 余项估计下列近似公式的误差 :}
\subsection{$\ln (1+x) \approx x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{6},$ 当 $0 \leqslant x \leqslant 1$}
\textbf{解}\quad
$$
	\begin{array}{c}
		\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{6}+\frac{x^{7}}{7(1+\xi x)^{7}}, \quad 0 \leqslant \xi \leqslant 1 \\
		R_{n}(x)=\left|\frac{x^{7}}{7(1+\xi x)^{7}}\right| \leqslant \frac{1}{7} \approx 0.1429
	\end{array}
$$

\subsection{$\cos x \approx 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720},$ 当 $|x|<1$}
\textbf{解}\quad

$$
	\begin{array}{l}
		\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\frac{\cos (\theta x)}{8 !} x^{8} \\
		R_{n}(x)\left|\frac{\cos (\theta x) x^{8}}{8 !}\right| \leqslant \frac{x^{8}}{8 !} \leqslant \frac{1}{8 !} \approx 2.42 \times 10^{-5}
	\end{array}
$$

\section{设 $f(x)$ 在 [0,1] 上有二阶导数 $,|f(x)| \leqslant a,\left|f^{\prime \prime}(x)\right| \leqslant b,$ 其中 $a, b$ 是非负实数,求证: 对一切 $c \in(0,1)$ 有 $\left|f^{\prime}(c)\right| \leqslant 2 a+\frac{b}{2}$.}
\textbf{证}\quad
在$c$处展开

$$
	\begin{aligned}
		f(0) & =f(c)+f^{\prime}(c)(-c)+\frac{f^{\prime \prime}\left(\xi_{1}\right)}{2} c^{2}, \quad \xi_{1} \in(0, c)
		\\f(1)&=f(c)+f^{\prime}(c)(1-c)+\frac{f^{\prime \prime}\left(\xi_{2}\right)}{2}(1-c)^{2}, \quad \xi_{2} \in(c, 1)
	\end{aligned}
$$

二者相减:

$$
	f(1)-f(0)=f^{\prime}(c)+\frac{1}{2}(1-c)^{2} f^{\prime \prime}\left(\xi_{2}\right)+\frac{1}{2} c^{2} f^{\prime \prime}\left(\xi_{1}\right)
$$

$$
	\begin{aligned}\left|f^{\prime}(c)\right| & \leqslant|f(1)|+|f(0)|+\frac{1}{2}(1-c)^{2}\left|f^{\prime \prime}\left(\xi_{2}\right)\right|+\frac{1}{2} x^{2}\left|f^{\prime \prime}\left(\xi_{1}\right)\right| \\
		                           & \leqslant 2 a+\frac{1}{2}\left[(1-c)^{2}+c^{2}\right] b                                                                                                           \\
		                           & \leqslant 2 a+\frac{1}{2}[(1-c)+c] b=2 a+\frac{b}{2}
	\end{aligned}
$$

\section{设 $f(x)$ 在$\mathbb{R}$上二次可微,且 $\forall x \in \mathbb{R}$,有$|f(x)| \leqslant M_{0},\left|f^{\prime \prime}(x)\right| \leqslant M_{2}$}
\subsection{写出 $f(x+h), f(x-h)$ 关于 $h$ 的带拉格朗日余项的泰勒公式}
\textbf{证}\quad
$$f(x+h)=f(x)+f^{\prime}(x) h+\frac{f^{\prime \prime}\left(x+\theta_{1} h\right)}{2} h^{2} \quad\left(0<\theta_{1}<1\right)$$
$$f(x-h)=f(x)-f^{\prime}(x) h+\frac{f^{\prime \prime}\left(x-\theta_{2} h\right)}{2} h^{2} \quad\left(0<\theta_{2}<1\right)$$


\subsection{求证 : 对 $\forall h>0,$ 有 $\left|f^{\prime}(x)\right| \leqslant \frac{M_{0}}{h}+\frac{h}{2} M_{2}$}
\textbf{证}\quad
将第12.1得到的两个泰勒公式相减,得
$$
2 f^{\prime}(x) h=f(x+h)-f(x-h)+\frac{f^{\prime \prime}\left(x-\theta_{2} h\right)}{2} h^{2}-\frac{f^{\prime \prime}\left(x+\theta_{1} h\right)}{2} h^{2}
$$

由此,利用条件 $|f(x)| \leqslant M_{0},\left|f^{\prime \prime}(x)\right| \leqslant M_{2},$ 即得
$$
\left|f^{\prime}(x)\right| \leqslant \frac{M_{0}}{h}+\frac{h}{2} M_{2}
$$


\subsection{求证 $:\left|f^{\prime}(x)\right| \leqslant \sqrt{2 M_{0} M_{2}}$}
\textbf{证}\quad
设 $g(h) = \frac{M_{0}}{h}+\frac{h}{2} M_{2},$ 则有
$$
g(h) \geqslant 2 \sqrt{\frac{M_{0}}{h} \cdot \frac{h}{2} M_{2}}=2 \sqrt{\frac{M_{0} M_{2}}{2} }=\sqrt{2M_0M_2}
$$

其中等号当 $\frac{M_{0}}{h}=\frac{h}{2} M_{2}$ 时, 即当 $h=\sqrt{\frac{2 M_{0}}{M_{2}}}$ 时成立. 将此 $h$值代入上式,即得

$$
\left|f^{\prime}(x)\right| \leqslant \sqrt{\frac{M_{0} M_{2}}{2}}+\sqrt{\frac{M_{0} M_{2}}{2}}=\sqrt{2 M_{0} M_{2}}
$$

\section{设 $f(x)$ 在 $[0, a]$ 上二阶可导,且 $\left|f^{\prime \prime}(x)\right| \leqslant M, f(x)$ 在 $(0, a)$ 上取得最大值,则$\left|f^{\prime}(0)\right|+\left|f^{\prime}(a)\right| \leqslant M a$}
\textbf{证}\quad
把$f^{\prime}(x)$在最大值点$c$用带拉格朗日余项的泰勒公式展开

$$f^{\prime}(a)=f^{\prime}(c)+f^{\prime\prime}(\xi_1)(a-c)$$
$$f^{\prime}(0)=f^{\prime}(c)+f^{\prime \prime}(\xi_2)(0-c)$$

$$\begin{aligned}\left|f^{\prime}(a)\right|+\left|f^{\prime}(0)\right|&=\left|f^{\prime \prime}\left(\xi_{1}\right)(a-c)\right|+\left|f^{\prime\prime}\left(\xi_{2}\right) \cdot c\right| \\  &\leq M a \end{aligned}$$

\section{设函数 $f(x)$ 在 $[a, b]$ 上有一阶连续导数,在 $(a, b)$ 内二阶可导,且有 $f(a)=f(b)=0$, 证明:任取 $x \in(a, b),$ 存在 $\xi \in(a, b),$ 使得 $f(x)=\frac{f^{\prime \prime}(\xi)}{2}(x-a)(x-b)$}
\textbf{证}$1^{\circ}$\quad
固定 $x \in(a, b),$ 令 $\lambda=\frac{2 f(x)}{(x-a)(x-b)},$ 则题目等价于证明存在 $\xi \in(a, b),$ 使得
$f^{\prime \prime}(\xi)=\lambda .$ 构造辅助函数 $g(t)=f(t)-\frac{\lambda}{2}(t-a)(t-b),$ 则 $g(t)$ 在 $[a, b]$ 上连续,在 $(a, b)$ 内可导,并且 $g(a)=g(b)=0,$ 从 $\lambda=\frac{2 f(x)}{(x-a)(x-b)}$ 的定义可知 $g(x)=0 .$ 在区间 $[a, x]$, $[x, b]$ 上分别对 $g(t)$ 用罗尔中值定理,可知,存在 $\xi_{1} \in(a, x), \xi_{2} \in(x, b),$ 使得 $g^{\prime}\left(\xi_{1}\right)=g^{\prime}\left(\xi_{2}\right)=0 .$ 然后再在区间 $\left[\xi_{1}, \xi_{2}\right]$ 上对 $g^{\prime}(t)$ 用罗尔中值定 理, 则存在 $\xi \in\left(\xi_{1}, \xi_{2}\right),$ 使得
$g^{\prime \prime}(\xi)=0,$ 即
$$
g^{\prime}(t)=f^{\prime}(t)-\frac{\lambda}{2}[(t-a)+(t-b)], \quad g^{\prime \prime}(t)=f^{\prime \prime}(t)-\lambda
$$

所以 $g^{\prime \prime}(\xi)=f^{\prime \prime}(\xi)-\lambda=0 \Leftrightarrow f^{\prime \prime}(\xi)=\lambda=\frac{2 f(x)}{(x-a)(x-b)},$ 即 $f(x)=\frac{f^{\prime \prime}(\xi)}{2}(x-a)(x-b)$

\textbf{证}$2^{\circ}$\quad
将要证明的不等式改写为 $f^{\prime \prime}(\xi)=\frac{2 f(x)}{(x-a)(x-b)},$ 记 $g(x)=(x-a)(x-b)$
则由已知条件和 $g(x)$ 的定义可知 $: f(a)=f(b)=g(a)=g(b)=0,$ 又因为涉及二阶导数,所以要使用两次中值定理,取中点 $c=\frac{a+b}{2},$ 将区间 $[a, b]$ 分为两部分 $\left[a, \frac{a+b}{2}\right],\left[\frac{a+b}{2}, b\right] .$ 如果 $f^{\prime}(c)=\frac{a+b}{2} \neq 0,$ 则 $\frac{2 f(x)}{(x-a)(x-b)}$ 中分子和分母的导数不会同时为零,因此可以 利用函数 $f(a)=f(b)=g(a)=g(b)=0,$ 在区间 $\left[a, \frac{a+b}{2}\right],\left[\frac{a+b}{2}, b\right]$ 上利用 $\mathrm{Cauchy}$ 中
值定理,则
$$
\frac{2 f(x)}{(x-a)(x-b)}=\frac{f^{\prime}\left(\xi_{1}\right)}{\xi_{1}-c}=\frac{f^{\prime}\left(\xi_{2}\right)}{\xi_{2}-c}=\frac{f^{\prime}\left(\xi_{1}\right)-f^{\prime}\left(\xi_{2}\right)}{\xi_{1}-\xi_{2}}
$$

其中, $\xi_{1} \in(a, x), \xi_{2} \in(x, b),$ 因为函数 $f(x)$ 在 $\left[\xi_{1}, \xi_{2}\right]$ 上二阶可导, 所以在 $\left[\xi_{1}, \xi_{2}\right]$ 上对 $f^{\prime}(x)$ 使用 $\mathrm{Lagrange}$ 中值定理,则存在 $\xi \in\left(\xi_{1}, \xi_{2}\right) \subset(a, b),$ 使得
$$
f^{\prime \prime}(\xi)=\frac{f^{\prime}\left(\xi_{1}\right)-f^{\prime}\left(\xi_{2}\right)}{\xi_{1}-\xi_{2}}=\frac{2 f(x)}{(x-a)(x-b)}
$$

\section{设函数 $f(x)$ 在 $[a, b]$ 上有二阶连续导数,在 $(a, b)$ 上三阶可导,且有 $f(a)=f^{\prime}(a)=f(b)=0,$ 证明 : 任取 $x \in(a, b),$ 存在 $\xi \in(a, b),$ 使得$
f(x)=\frac{f^{\prime \prime \prime}(\xi)}{3 !}(x-a)^{2}(x-b)$}
\textbf{证}\quad
固定 $x \in(a, b),$ 令 $\lambda=\frac{6 f(x)}{(x-a)^{2}(x-b)},$ 则题目等价于证明存在 $\xi \in(a, b),$ 使得 $f^{\prime \prime \prime}(\xi)=\lambda$
构造辅助函数 $g(t)=f(t)-\frac{\lambda}{6}(t-a)^{2}(t-b),$ 则 $g(t)$ 在 $[a, b]$ 上连续,在 $(a, b)$ 内可导,并且 $g(a)=g(b)=0.$ 

从 $\lambda=\frac{2 f(x)}{(x-a)(x-b)}$ 的定义可知 $g(x)=0 .$ 在区间 $[a, x],[x, b]$ 上分 别对 $g(t)$ 用罗尔中值定理,可知,存在 $\xi_{1} \in(a, x), \xi_{2} \in(x, b),$ 使得 $g^{\prime}\left(\xi_{1}\right)=g^{\prime}\left(\xi_{2}\right)=0 .$ 

另一方面 $, g^{\prime}(t)=f^{\prime}(t)-\frac{\lambda}{6}\left[2(t-a)(t-b)+(t-a)^{2}\right],$ 显然 $g^{\prime}(a)=f^{\prime}(a)=0 .$ 因为 $g^{\prime}(a)=$
$g^{\prime}\left(\xi_{1}\right)=g^{\prime}\left(\xi_{2}\right)=0,$ 在区间 $\left[a, \xi_{1}\right],\left[\xi_{1}, \xi_{2}\right]$ 上对 $g^{\prime}(t)$ 用罗尔中值定理,则存在 $\eta_{1} \in\left(a, \xi_{1}\right),$
$\eta_{2} \in\left(\xi_{1}, \xi_{2}\right),$ 使得 $g^{\prime \prime}\left(\eta_{1}\right)=g^{\prime \prime}\left(\eta_{2}\right)=0 .$ 因为 $f(x)$ 在 $\left[\eta_{1}, \eta_{2}\right]$ 上三阶可导,所以 $g^{\prime \prime}(t)$ 在
$\left[\eta_{1}, \eta_{2}\right]$上可导,利用罗尔中值定理,可知存在 $\xi \in\left(\eta_{1}, \eta_{2}\right) \subset(a, b),$ 使得 $g^{\prime \prime \prime}(\xi)=0 .$
$$
g^{\prime \prime}(t)=f^{\prime \prime}(t)-\frac{\lambda}{6}[2(t-b)+2(t-a)+2(t-a)], \quad g^{\prime \prime \prime}(t)=f^{\prime \prime \prime}(t)-\lambda
$$

所以 $g^{\prime \prime \prime}(\xi)=f^{\prime \prime \prime}(\xi)-\lambda=0, f^{\prime \prime \prime}(\xi)=\lambda=\frac{6 f(x)}{(x-a)^{2}(x-b)},$ 即
$$
f(x)=\frac{f^{\prime \prime \prime}(\xi)}{3 !}(x-a)^{2}(x-b)
$$

\section{若函数 $f(x)$ 在 $[-1,1]$ 上三阶可导,且有 $f(0)=f^{\prime}(0)=0, f(1)=1, f(-1)=0 .$ 证明 : 存在 $\xi \in(-1,1),$ 使得 $f^{\prime \prime \prime}(\xi) \geqslant 3 .$}
\textbf{证}\quad
函数 $f(x)$ 在 $[-1,1]$ 上三阶可导,且 $x=0$ 为函数的一个驻点,把函数在 $x=0$ 处 三阶 $\mathrm{Taylor}$ 展开,有 $f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}\left(\eta_{1}\right)}{3 !} x^{3}=\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}\left(\eta_{1}\right)}{3 !} x^{3}$
$\eta_{1}$ 介于 $x, 0$ 之间. 把 $x=1, x=-1$ 代入,可得

$$
\begin{array}{l}
f(1)=\frac{f^{\prime \prime}(0)}{2 !}+\frac{f^{\prime \prime \prime}\left(\eta_{1}\right)}{3 !}=1, \quad \eta_{1} \in(0,1) \\
f(-1)=\frac{f^{\prime \prime}(0)}{2 !}-\frac{f^{\prime \prime \prime}\left(\eta_{2}\right)}{3 !}=0, \quad \eta_{2} \in(-1,0)
\end{array}
$$

两式相减,可得$f^{\prime \prime \prime}\left(\eta_{1}\right)+f^{\prime \prime\prime}\left(\eta_{2}\right)=6$

所以存在$\xi=max\{\eta_1,\eta_2\}$满足题目条件.

\section{设函数 $f(x)$ 在 $[a, b]$ 上有一阶连续导数,在 $(a, b)$ 上二阶可导,证明 :存在 $\xi \in(a, b),$
使得$f(a)-2 f\left(\frac{a+b}{2}\right)+f(b)=\frac{1}{4}(b-a)^{2} f^{\prime \prime}(\xi)$}
\textbf{证}\quad
$$f(a)=f\left(\frac{a+b}{2}\right)+f^{\prime}\left(\frac{a+b}{2}\right) \frac{a-b}{2}+\frac{f^{\prime \prime}\left(\xi_{1}\right)}{2}\left(\frac{a-b}{2}\right)^{2}, \quad a<\xi_{1}<\frac{a+b}{2}$$
$$f(b)=f\left(\frac{a+b}{2}\right)+f^{\prime}\left(\frac{a+b}{2}\right) \frac{b-a}{2}+\frac{f^{\prime \prime}\left(\xi_{2}\right)}{2}\left(\frac{b-a}{2}\right)^{2}, \quad \frac{a+b}{2}<\xi_{2}<b$$

两式相加可得
$$
f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{(b-a)^{2}}{4} \frac{f^{\prime \prime}\left(\xi_{1}\right)+f^{\prime \prime}\left(\xi_{2}\right)}{2}
$$

不妨设 $f^{\prime \prime}\left(\xi_{1}\right) \leqslant f^{\prime \prime}\left(\xi_{2}\right),$ 则
$$
f^{\prime \prime}\left(\xi_{1}\right) \leqslant \frac{f^{\prime \prime}\left(\xi_{1}\right)+f^{\prime \prime}\left(\xi_{2}\right)}{2} \leqslant f^{\prime \prime}\left(\xi_{2}\right)
$$

在 $\xi_{1}, \xi_{2}$ 之间,对 $f^{\prime \prime}(x)$ 应用达布定理,存在 $\xi$ 介于 $\xi_{1}, \xi_{2}$ 之间,使得
$$
f^{\prime \prime}(\xi) \leqslant \frac{f^{\prime \prime}\left(\xi_{1}\right)+f^{\prime \prime}\left(\xi_{2}\right)}{2}
$$

从而证明了 $$f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{(b-a)^{2}}{4} f^{\prime \prime}(\xi)$$

\section{设函数 $f(x)$ 在 $x_{0}$ 的某个邻域内有连续的 $n$ 阶导数,在 $x_{0}$ 点处有 $n+1$ 阶导数,且 $f^{(n+1)}\left(x_{0}\right) \neq 0 .$ 由 带 $\mathrm{Lagrange}$ 余项 的 Taylor 公式我们知道:对足够靠近 $x_{0}$ 的 $x,$ 存在 $\theta_{n}(x) \in(0,1),$ 使得$f(x)=\sum_{i=0}^{n-1} \frac{f^{(i)}\left(x_{0}\right)}{i !}\left(x-x_{0}\right)^{i}+\frac{f^{(n)}\left(x_{0}+\theta_{n}(x)\left(x-x_{0}\right)\right)}{n !}\left(x-x_{0}\right)^{n}$.证明:$\lim _{x \rightarrow x_{0}} \theta_{n}(x)=\frac{1}{n+1}$}
\textbf{证}\quad
按题设,我们有
$$
f(x+h)=f(x)+h f^{\prime}(x)+\cdots+\frac{h^{n}}{n !} f^{(n)}(x+\theta h)
$$

其中 $0<\theta<1 .$ 又因 $f^{(n+1)}(x)$ 存在,故
$$
f(x+h)=f(x)+h f^{\prime}(x)+\cdots+\frac{h^{n}}{n !} f^{(n)}(x)+\frac{h^{n+1}}{(n+1) !} f^{(n+1)}(x)+o\left(h^{n+1}\right)
$$

比较上面两式,得
$$
\frac{h^{n}}{n !} f^{(n)}(x+\theta h)=\frac{h^{n}}{n !} f^{(n)}(x)+\frac{h^{n+1}}{(n+1) !} f^{(n+1)}(x)+o\left(h^{n+1}\right)
$$

从而有
$$
\theta \cdot \frac{f^{(n)}(x+\theta h)-f^{(n)}(x)}{\theta h}=\frac{1}{n+1} f^{(n+1)}(x)+n ! \frac{o\left(h^{n+1}\right)}{h^{n+1}}
$$

由于 $\lim _{h \rightarrow 0} \frac{f^{(n)}(x+\theta h)-f^{(n)}(x)}{\theta h}=f^{(n+1)}(x) \neq 0,$ 故由上式知 $\lim _{h \rightarrow 0} \theta$ 存在,并且
$$
\lim _{h \rightarrow 0} \theta=\frac{\frac{1}{n+1} f^{(n+1)}(x)}{f^{(n+1)}(x)}=\frac{1}{n+1}
$$

\section{微分中值定理与微分方程}
\subsection{若形式为$f^{\prime}(\xi)+g(\xi)f(\xi)=0$}
考虑$F(x)=e^{\int g(x)dx}f(x)$

\subsection{若形式为$f^{\prime\prime}(\xi)+af^{\prime}(\xi)+bf(\xi)$}
考虑$F(x)=e^{\alpha x}(f^{\prime}(x)+\beta f(x))$

其中$\alpha + \beta = a, \alpha \beta = b$


\end{document}
